# Stagnation pressure on an accelerating surface

The usual stagnation pressure $$P_0$$ on an object (e.g., a sphere) moving at velocity $$U$$ is simply

$$P_0 = P_\infty + \frac{\rho}{2}U^2$$

where $$P_\infty$$ is the pressure of the free stream.

If the object is accelerating, thus having a time dependent velocity $$U(t)$$, is the stagnation pressure given by the following equation?

$$P_0 = P_\infty + \frac{\rho}{2}U(t)^2$$

Or must we use the unsteady Bernoulli equation?

The expression for stagnation pressure is derived directly from Bernoulli's principle. Let's go through the derivation of Bernoulli's principle to understand the assumptions involved.

Consider a pipe whose axis is along the x-axis of the cartesian system. Let $$A$$ be its cross-sectional area and a fluid of density $$\rho$$ flow through the pipe. We will concentrate on a fluid parcel of length $$dx$$. Its volume is $$Adx$$ and hence, mass of the fluid parcel would be $$\rho Adx$$. Also, let $$dP$$ be the change in the pressure on moving a distance of $$dx$$ and $$\frac{dx}{dt}$$ be the flow velocity.

Ignoring gravity, for the time being, the only force acting on the fluid parcel is due to the pressure difference $$dP$$, $$F = -AdP$$ (negative sign occurs since the fluid moves towards regions of lower pressure). Applying Newton's second law,

$$\rho Adx \frac{dv}{dt} = -AdP$$ $$\rho \frac{dv}{dt} = -\frac{dP}{dx}$$

Here comes another assumption: the flow velocity is steady, i.e., $$v \equiv v(x)$$ and the time dependence is not explicit; but comes in because the fluid parcel is moving and the velocity is a function of position, $$v \equiv v(x(t))$$. Therefore,

$$\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}$$ $$= \frac{dv}{dx}v = \frac{d}{dx}(\frac{1}{2}v^2 )$$

Now assuming fluid to be incompressible (constant $$\rho$$),

$$\frac{d}{dx}(\rho \frac{v^2}{2} + p) = 0$$

Integrating with respect to x,

$$\rho \frac{v^2}{2} + p = constant$$

If the fluid were to accelerate then the velocity field would be an explicit function of time, $$v \equiv v(x(t),t)$$ and $$\frac{dv}{dt} = v\frac{dv}{dx} + \frac{\partial v}{\partial t}$$ and we will not be able to proceed with the same derivation.

So, you probably should use the unsteady Bernoulli equation.